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NEW QUESTION: 1
Non-Discretionary Access Control. A central authority determines what subjects can have access to certain objects based on the organizational security policy. The access controls may be based on?
A. The individual's role in the organization.
B. The group-dynamics as they relate to the individual's role in the organization.
C. The societies role in the organization.
D. The group-dynamics as they relate to the master-slave role in the organization.
Answer: A
Explanation:
An access control model defines a computer and/or network system's rules for user access to information resources. Access control models provide confidentiality, integrity and also provide accountability through audit trails. An audit trail documents the access of an object by a subject with a record of what operations were performed.
Operations include: read, write, execute and own.
Non-Discretionary Access Control is usually role-based, centrally administered with authorization decisions based on the roles individuals have within an organization (e.g.
bank teller, loan officer, etc. in a banking model). A system's security administrator grants and/or revokes system privileges based on a user's role. This model works well for corporations with a large turnover of personnel.

NEW QUESTION: 2
What type of Output variable do all Get Mail activities return? (POP3, IMAP, Outlook, Exchange) Options are :
A. MailMessage
B. List
C. List
Answer: B

NEW QUESTION: 3
What do you use to activate SAP Fiori search and SAP Fiori fact sheets?
Please choose the correct answer. Response:
A. Business functions
B. Program exit
C. BAdl
D. User exit
Answer: A

NEW QUESTION: 4
customers.txtにアクセス可能であり、複数の行が含まれていると仮定します。
customers.txtファイルの内容を印刷するコードフラグメントはどれですか？
A. Stream <String> lines = Files.lines（Paths.get（ "customers.txt"））; lines.forEach（c）-> System.out.println（c））;
B. Stream <Path> stream = Files.find（Paths.get（ "customers.txt"））; stream.forEach（c）-> System.out.println（c））;
C. Stream <Path> stream = Files.list（Paths.get（ "customers.txt"））; stream.forEach（c）-> System.out.println（c））;
D. Stream <String> stream = Files.find（Paths.get（ "customers.txt"））; stream.forEach（（String c）-> System.out.println（c））;
Answer: D